2
$\begingroup$

I can find this using the fact that $\sin(\sin^{-1}(x)) = x$, for all $x\in[-1,1].$

Now, differentiate.

$$\frac{d}{d\sin^{-1}(x)}\sin(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x)= \frac{d}{dx} x= 1$$

$$\cos(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x) = 1$$

$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))}$$

$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-\sin^2(\sin^{-1}(x))}}$$

$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$$

However, what if I wanted to differentiate this like $\ \sin^{-1}(\sin(x))$ without knowing the fact that $\ \frac{d}{dx}\sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$ ? Is there a solution for it? I keep getting stuck at a certain step when I try this...

$\endgroup$
1
$\begingroup$

let $ y = \arcsin(\sin x) $. All you have to do then is consider $$ \sin y = \sin(\arcsin(\sin x))$$ and differentiate implicitly.

$\endgroup$
  • $\begingroup$ Could you show me how this would work? Thanks! $\endgroup$ – user13123 Nov 23 '17 at 19:32
4
$\begingroup$

If you're asking how to differentiate$$y=\arcsin(\sin x)$$without using the fact that$$\frac {d}{dx}\,\arcsin x=\frac 1{\sqrt{1-x^2}}$$Then the easiest way is to use implicit differentiation. Setting your expression equal to $y$, we have that$$\sin y=\sin x$$So$$y_x\cdot\cos y=\cos x\quad\implies\quad y_x=\frac {\cos x}{\cos y}$$

$\endgroup$
  • $\begingroup$ How would this work with $sin^{-1}x$? Do I need to set $sin^{-1}x = sin^{-1}y$? $\endgroup$ – user13123 Nov 23 '17 at 5:02
  • $\begingroup$ @user13123 I'm sorry, but I'm slightly confused what you're asking. If you're asked to differentiate a trigonometric inverse function, the first thing that comes to mind should be implicit differentiation. $\endgroup$ – Crescendo Nov 23 '17 at 5:23
  • $\begingroup$ Ok... but how would you implicitly differentiate $sin^{-1}x(sinx)$? $\endgroup$ – user13123 Nov 23 '17 at 5:39
  • $\begingroup$ @user13123 Well if we let $y=\arcsin(x\sin x)$ then$$\sin y=x\sin x$$ $\endgroup$ – Crescendo Nov 23 '17 at 17:45
1
$\begingroup$

You could do it like this: first, define $f$ by $f(x) = \sin^{-1}x,$ where $-1 \leq x \leq 1$ and $-\frac\pi2 \leq \sin^{-1}x \leq \frac\pi2.$ (Recall that $\sin^{-1}$ is not a real function for $|x| > 1$ and that $\sin$ is not a one-to-one function, so we have to choose an appropriate domain and range of $\sin^{-1}$; these are the usual choices.) Then for $-\frac\pi2 \leq \theta \leq \frac\pi2,$ $$ f(\sin \theta) = \theta. \tag1 $$ Take the derivative of each side of $(1)$ with respect to $\theta$: $$ \frac{d}{d\theta}\left(f(\sin\theta)\right) = 1. \tag2 $$ Evaluate the left-hand side of $(2)$ using the chain rule: $$ \frac{d}{d\theta}\left(f(\sin\theta)\right) = f'(\sin\theta)\frac{d}{d\theta}\left(\sin\theta\right) = f'(\sin\theta) \cos\theta. \tag3 $$ Combine $(2)$ and $(3)$: $$ f'(\sin\theta) \cos\theta = 1. \tag4 $$ Divide by $\cos\theta$ on both sides of $(4)$: $$ f'(\sin\theta) = \frac{1}{\cos\theta}. \tag5 $$ Let $x = \sin\theta.$ Then $1 - x^2 = \cos^2\theta,$ and since $-\frac\pi2 \leq \theta \leq \frac\pi2,$ it follows that $\cos\theta\geq 0,$ so we have $\cos\theta = \sqrt{1 - x^2}.$ Making these substitutions in $(5)$, $$ f'(x) = \frac{1}{\sqrt{1 - x^2}}. $$ But $f'(x)=\frac{d}{dx}\sin^{-1}x,$ so $$ \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.